Percentages — Paper 4 · IGCSE Mathematics (9-1) 0980

1 (a) The price of a newspaper increased from $0.97 to $1.13 . Calculate the percentage increase. … % [3] (b) One day, the newspaper had 60 pages of news and advertisements. The ratio number of pages of news : number of pages of advertisements = 5 : 7. (i) Calculate the number of pages of advertisements. … [2] (ii) Write the number of pages of advertisements as a percentage of the number of pages of news. … % [1] (c) On holiday Maria paid 2.25 euros for the newspaper when the exchange rate was $1 = 0.9416 euros. At home Maria paid $1.13 for the newspaper. Calculate the difference in price. Give your answer in dollars, correct to the nearest cent. $ … [3] (d) The number of newspapers sold decreases exponentially by x% each year. Over a period of 21 years the number of newspapers sold decreases from 1 763 000 to 58 000. Calculate the value of x. x = … [3] (e) Every page of the newspaper is a rectangle measuring 43 cm by 28 cm, both correct to the nearest centimetre. Calculate the upper bound of the area of a page. … cm2 [2]

14 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) 16.5 or 16.49... 3 M2 for 1.13 − 0.97[× 100] oe or 1.13 × 100 oe 0.97 0.97 1.13 or M1 for oe 0.97 1(b)(i) 35 2 M1 for 60 ÷ ( 5 + 7 ) 1(b)(ii) 140 1 1(c) $1.26 final answer 3 B2 for 1.259... or 1.26 but not as final answer or M1 for 2.25 ÷ 0.9416 If 0 scored, SC1 for 1.13 × 0.9416 1(d) 15[.0…] 3 58000 M2 for 21 oe 1763000 or M1 for 58000 = 1763000 ( k ) 21 1(e) 1239.75 2 B1 for 43 + 0.5 or 28 + 0.5 oe seen

This question in 0980/42 May/June 2019

2 (a) Ali and Mo share a sum of money in the ratio Ali : Mo = 9 : 7. Ali receives $600 more than Mo. Calculate how much each receives. Ali $ … Mo $ … [3] (b) In a sale, Ali buys a television for $195.80 . The original price was $220. Calculate the percentage reduction on the original price. … % [3] (c) In the sale, Mo buys a jacket for $63. The original price was reduced by 25%. Calculate the original price of the jacket. $ … [3]

9 marks

Mark scheme: 2(a) [Ali] 2700 3 B2 for one correct or for correct values [Mo] 2100 reversed or M1 for 600 ÷ (9 – 7) or for any equation that would lead to an answer of 300, 2700 or 2100, or 4800 (for the total) 2(b) 11 3 220 − 1958. M2 for [× 100] or for 220 1958. [100 − ] × 100 220 195 8. or M1 for 220 – 195.8 or for or a 220 correct implicit equation for percentage 195.8 − 220 reduction or for 220 2(c) 84 3 63 M2 for oe 25 1 − 100 or M1 for associating 63 with (100 – 25)% or a correct implicit equation for the original price.

This question in 0980/41 Oct/Nov 2019

1 (a) (i) Divide $24 in the ratio 7 : 5. $ … , $ … [2] (ii) Write $24.60 as a fraction of $2870. Give your answer in its lowest terms. … [2] (iii) Write $1.92 as a percentage of $1.60 . … % [1] (b) In a sale the original prices are reduced by 15%. (i) Calculate the sale price of a book that has an original price of $12. $ … [2] (ii) Calculate the original price of a jacket that has a sale price of $38.25 . $ … [2] (c) (i) Dean invests $500 for 10 years at a rate of 1.7% per year simple interest. Calculate the total interest earned during the 10 years. $ … [2] (ii) Ollie invests $200 at a rate of 0.0035% per day compound interest. Calculate the value of Ollie’s investment at the end of 1 year. [1 year = 365 days.] $ … [2] (iii) Edna invests $500 at a rate of r % per year compound interest. At the end of 6 years, the value of Edna’s investment is $559.78 . Find the value of r. r = … [3]

16 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 14, 10 2 M1 for 24 ÷ (7 + 5) 1(a)(ii) 3 2 B1 for correct fraction not in lowest terms 350 1(a)(iii) 120 1 1(b)(i) 10.2[0] 2 15 M1 for × 12 oe or better 100 1(b)(ii) 45 2 38.25 M1 for oe 15 1 − 100 1(c)(i) 85 2 500 × 1.7 × 10 M1 for oe 100 1(c)(ii) 203 or 202.5 to 202.6 2 365  0.0035  M1 for 200 ×  1 +   100  1(c)(iii) 1.9 3 559.78 M2 for 6 500  r 6 or M1 for 500  1 +  = 559.78  100 

This question in 0980/42 May/June 2020

9 (a) There are 32 students in a class. 5 do not study any languages. 15 study German (G). 18 study Spanish (S). G S (i) Complete the Venn diagram to show this information. [2] (ii) A student is chosen at random. Find the probability that the student studies Spanish but not German. … [1] (iii) A student who studies German is chosen at random. Find the probability that this student also studies Spanish. … [1] (b) A bag contains 54 red marbles and some blue marbles. 36% of the marbles in the bag are red. Find the number of blue marbles in the bag. … [2] (c) Another bag contains 15 red beads and 10 yellow beads. Ariana picks a bead at random, records its colour and replaces it in the bag. She then picks another bead at random. (i) Find the probability that she picks two red beads. … [2] (ii) Find the probability that she does not pick two red beads. … [1] (d) A box contains 15 red pencils, 8 yellow pencils and 2 green pencils. Two pencils are picked at random without replacement. Find the probability that at least one pencil is red. … [3]

12 marks

Mark scheme: 9(a)(i) 2 B1 for two correct values 5 Or 9 6 12 B1 5 outside and total in G = 15 and total in G S S = 18 9(a)(ii) 3 1 their 12 oe FT 8 32 9(a)(iii) 2 1 their 6 oe FT 5 15 9(b) 96 2 36 54 54 M1 for = oe or 36 = × 100 64 x ( 54 + b ) oe If 0 scored SC1 for answer 150 9(c)(i) 9 2 15 15 oe M1 for × oe 25 25 25 9(c)(ii) 16 1 FT 1 – their (c)(i) oe 25 9(d) 17 3 10 9 oe M2 for 1 − × oe 20 25 24 15 14 15 8 15 2 8 15 or for × + × + × + × 25 24 25 24 25 24 25 24 2 15 + × oe 25 24 or M1 for one correct relevant product

This question in 0980/41 Oct/Nov 2020

1 (a) A 2.5-litre tin of paint costs $13.50 . In a sale, the cost is reduced by 14%. (i) Work out the sale price of this tin of paint. $ … [2] (ii) Work out the cost of buying 42.5 litres of paint at this sale price. $ … [2] (b) Henri buys some paint in the ratio red paint : white paint : green paint = 2 : 8 : 5. (i) Find the percentage of this paint that is white. … % [1] (ii) Henri buys a total of 22.5 litres of paint. Find the number of litres of green paint he buys. … litres [2] (c) Maria paints a rectangular wall. The length of the wall is 20.5 m and the height is 2.4 m, both correct to 1 decimal place. One litre of paint covers an area of exactly 10 m2. Calculate the smallest number of 2.5-litre tins of paint she will need to be sure all the wall is painted. Show all your working. … [4]

11 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 11.61 final answer 2  14  M1 for 13.5[0] ×  1 −  oe  100  or B1 for 1.89 1(a)(ii) 197.37 final answer 2 FT 17 × their (a)(i) exact or correct to nearest cent M1 for 42.5 ÷ 2.5 1(b)(i) 53.3 or 53.33… 1 1(b)(ii) 7.5 2 M1 for 22.5 ÷ (2 + 8 + 5) oe soi 1(c) 20.55 × 2.45 oe M2 M1 for 20.5 + 0.05 oe seen or 2.4 + 0.05 oe seen If 0 scored, SC1 here for 20.45 × 2.35 oe 3 nfww A2 M1 for their area ÷ 10 ÷ 2.5 oe

This question in 0980/42 May/June 2021

6 (a) At a festival, 380 people out of 500 people questioned say that they are camping. There are 55 300 people at the festival. Calculate an estimate of the total number of people camping at the festival. … [2] (b) 12 friends travel to the festival. 5 travel by car, 4 travel by bus and 3 travel by train. Two people are chosen at random from the 12 friends. Calculate the probability that they travel by different types of transport. … [4] (c) Arno buys a student ticket for $43.68 . This is a saving of 16% on the full price of a ticket. Calculate the full price of a ticket. $ … [2] (d) At a football match, there are 29 800 people, correct to the nearest 100. (i) At the end of the football match, the people leave at a rate of 400 people per minute, correct to the nearest 50 people. Calculate the lower bound for the number of minutes it takes for all the people to leave. … min [3] (ii) At a cricket match there are 27 500 people, correct to the nearest 100. Calculate the upper bound for the difference between the number of people at the football match and at the cricket match. … [2]

13 marks

Mark scheme: 6(a) 42 028 2 380 M1 for oe soi isw 500 6(b) 47 4 0.712[1…] oe 66  5 4   4 3   5 3  M3 for 2   2   2         12 11   12 11   12 11  oe  5 4 4 3 3 2  or 1 –      oe    12 11 12 11 12 11  or M2 for sum of 3 or more correct product pairs and no incorrect pairs 5 4 4 3 3 2 or for      and no other 12 11 12 11 12 11 pairs k j or M1 for  seen 12 11 94 If 0 scored SC1 for answer oe 144 6(c) 52 2 100  16 M1 for x   43.68 oe or better 100 6(d)(i) 70 or 70.16[5…] or 70.17 or 70.2 3 29750 to 29800 29750 to 29800 M2 for or or 400  25 400  24 29800  50 400to425 or B1 for 29 750 or 29 850 or 29 849 or 375 or 425 or 424 seen 6(d)(ii) 2399 2 B1 for 27 450 or 27 550 or 27 549 or 29 850 or or 2400 nfww 29 849 seen

This question in 0980/42 May/June 2022

1 (a) Calculate the volume of (i) a solid cylinder with radius 6 cm and height 14 cm, … cm3 [2] (ii) a solid hemisphere with radius 6 cm. 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3 … cm3 [2] (b) NOT TO SCALE 14 cm 6 cm The cylinder and hemisphere in part (a) are joined to form the solid in the diagram. The solid is made of steel and 1 cm 3 of steel has a mass of 7.85 g. (i) Show that 1 cm 3 of steel has a mass of 0.007 85 kg. [1] (ii) Calculate the total mass of the solid. … kg [2] (c) 2000 cm 3of iron is melted down and some of it is used to make 50 spheres with radius 2 cm. (i) Calculate the percentage of iron that is left over. 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3 … % [3] (ii) The iron left over is then made into a cube. Calculate the length of an edge of the cube. … cm [1] (d) A solid cone has radius 3R cm and slant height 9R cm. A solid cylinder has radius x cm and height 7x cm. The total surface area of the cone is equal to the total surface area of the cylinder. Given that R = kx , find the value of k. [The curved surface area, A, of a cone with radius r and slant height l is A = rrl .] k = … [4]

15 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 1580 or 1583 to 1584 2 M1 for π  6 2  14 1(a)(ii) 452 or 452.3 to 452.4... 2 3 M1 for 1 4  π  6  2 3 1(b)(i) 7.85 ÷ 1000 [= 0.00785] M1 1(b)(ii) 16[.0] or 15.95 to 15.99 2 FT {their (a)(i) + their (a)(ii)}  0.00785 evaluated to 3 sig fig or better M1 for (their (a)(i) + their (a)(ii)) × 0.00785 1(c)(i) 16.2 or 16.21 to 16.23 3 4 3 2000 − 50  π 2 3 M2 for 100  2000 4 3 50  π 2 3 or for  100 2000 4 3 50  π 2 3 or M1 for 2000 1(c)(ii) 6.87 or 6.870 to 6.872 1  4 3  FT 3 2000 − their 50  π 2    3  evaluated to 3sf or better 1(d) 2 4 M1 for [π](3 R ) 2 + [π]3 R  9 R oe oe 3 M1 for 2[π]x 2 + 2[π]x  7 x oe M1 for their area of cone = their area of cylinder seen

This question in 0980/41 Oct/Nov 2022

2 (a) Anil changes $830 into euros when the exchange rate is 1 euro = $1.16 . He spends 500 euros. He then changes the remaining money back into dollars at the same exchange rate. Work out how much, in dollars, Anil receives. $ … [3] (b) In 2021, Anil earns $37 000. (i) He spends $12 400 on bills in 2021. Calculate the percentage of his earnings he spends on bills. … % [2] (ii) His earnings of $37 000 increase by 3.2% in 2022. Calculate his earnings in 2022. $ … [2] (c) Anil invests $3500 in an account that pays a rate of 2.4% per year compound interest. (i) Calculate the total interest earned at the end of 5 years. $ … [3] (ii) Find the number of complete years before Anil has at least $5000 in this account. … years [3]

13 marks

Mark scheme: 2(a) 249.98 to 250[.0…] 3 M2 for 830 – 500 × 1.16 or M1 for 500 × 1.16 OR M1 for 830 ÷ 1.16 M1 for (their 715.5… – 500 ) × 1.16 2(b)(i) 33.5 or 33.51… 2 12400 M1 for [ 100] oe 37000 If 0 scored, SC1 for answer 66.5 or 66.48 to 66.49 2(b)(ii) 38 184 cao 2  3.2  M1 for 37 000   1   oe  100  or B1 for 1184 2(c)(i) 441 or 440.6 3 B2 for answer 3941 or 3940.6 or 3940.64 or 440.64 to 440.65 to 3940.65  2.4  5 or M2 for 3500 ×  1   – 3500  100   2.4  5 or M1 for 3500 ×  1   oe isw  100  2(c)(ii) 16 3 B2 for 15[.0] nfww to 15.1  2.4 15 or M2 for 3500 ×  1   oe seen  100   2.4 16 or 3500 ×  1   oe seen  100  or M1 for  2.4  n (3500 or their 3941) ×  1    100  associated with 5000 oe

This question in 0980/42 May/June 2023

3 (a) The table shows information about some of the planets in the solar system. Planet Diameter (km) Average distance from the Sun (km) Earth 12 800 1.496 # 108 Mars 6 800 2.279 # 108 Jupiter 143 000 7.786 # 108 Saturn 120 500 1.434 # 109 Neptune 49 500 4.495 # 109 (i) The average distance of Mars from the Sun is 2.279 # 108 km . Write this distance as an ordinary number. … km [1] (ii) The planet Uranus has a diameter that is 35.8% of the diameter of Jupiter. Calculate the diameter of Uranus. … km [2] (iii) The ratio diameter of Neptune : diameter of Saturn can be written in the form 1 : n. Find the value of n. n = … [1] (iv) Find the average distance of Neptune from the Sun as a percentage of the average distance of the Earth from the Sun. … % [2] (v) Distances within the solar system are also measured in astronomical units (AU). The average distance of Jupiter from the Sun is 5.20 AU. Calculate the average distance of Mars from the Sun in astronomical units. … AU [2] (vi) The diameter of Mars is 39.2% greater than the diameter of Mercury. Calculate the diameter of Mercury. … km [2] (b) One light year is the distance that light travels in a year of 365.25 days. The speed of light is .29979 # 105 kilometres per second. (i) Show that one light year is 9.461 # 1012 km , correct to 4 significant figures. [2] (ii) The distance from the Andromeda Galaxy to Earth is 2.40 # 1019 km . Calculate the time taken for light to travel from this galaxy to Earth. Give your answer in millions of years. … million years [2]

14 marks

Mark scheme: 3(a)(i) 227 900 000 1 3(a)(ii) 51 200 or 51 190 or 51 194 2 35.8 M1 for  143 000 100 After 0 scored SC1 for answer figs 512 or figs 5119 or figs 51194 3(a)(iii) 2.43 or 2.434… 1 3(a)(iv) 3000 or 3004 to 3005 2 4.495  10 9 M1 for [ 100] oe 1.496  108 After 0 scored SC1 for answer figs 3 or figs 3004…. or figs 3005 3(a)(v) 1.52 or 1.522… 2 B1 for 1AU = 1.5[0]  108 or 1.497…  108 [km] or 1km = 6.68  10−9 or 6.678 10−9  AU OR 5.2  2.279[108 ] M1 for oe 7.786[108 ] After 0 scored SC1 for answer figs 152 or figs 1522…… 3(a)(vi) 4890 or 4885… 2  39.2  M1 for d   1 +  = 6800 oe  100  3(b)(i) 2.9979  105  602  24  M1 365.25 After M0 SC1 for 2.9979  105  31557600 oe = 9.4606…  1012 A1 3(b)(ii) 2.54 or 2.536 to 2.537 2 2.4  1019 M1 for 12 oe 9.461  10

This question in 0980/41 Oct/Nov 2023

1 (a) A fruit drink is made using 1.5 litres of apple juice and 450 millilitres of mango juice. Write the ratio apple juice : mango juice in its simplest form. … : … [2] (b) One litre of fruit drink is shared between three cups. The amount in the cups is in the ratio 9 : 6 : 10. Calculate the number of millilitres in each cup. … ml , … ml , … ml [3] (c) A shop buys bottles of the fruit drink for $3.20 each. It sells them at a profit of 15%. Calculate the selling price of each bottle of fruit drink. $ … [2] (d) The number of bottles of fruit drink sold has grown exponentially at a constant rate of 2.5% per year. 5 years ago, the shop sold 16 620 bottles. Calculate the number of bottles sold this year. … [2] (e) d cm NOT TO 23 cm SCALE 18.5 cm The bottles of juice are 18.5 cm tall, correct to the nearest millimetre. They are stored on shelves. The distance between the shelves is 23 cm, correct to the nearest centimetre. Calculate the lower bound for the distance, d cm, between the top of a bottle and the shelf above it. … cm [3]

12 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) 10 : 3 final answer 2 M1 for 1500 : 450 oe in ratio form If 0 scored SC1 for answer 3 : 10 1(b) 360 240 400 3 B2 for answer 0.36 0.24 0.4 or for answer two of 360 240 400 1000 or M1 for [ k ] where k = 1, 9, 9  6  10 6 or 10 If 0 scored, SC1 for answer with 3 values in ratio 9 : 6 : 10 in that order 1(c) 3.68 cao 2  15  M1 for  1    3.2 oe  100  or B1 for answer 0.48 1(d) 18 804[.0...] 2  2.5  5 1 for 16620   1   oe  100  1(e) 3.95 3 M2 for 22.5 – (18.5 to 18.6) or (22 to 23) −18.55 or M1 for 23 – 0.5 oe seen or 23 + 0.5 oe seen or 18.5– 0.05 oe seen or 18.5 + 0.05 oe seen

This question in 0980/42 May/June 2024

4 (a) Enzo, Rashid and Blessy each swim as many lengths of a swimming pool as they can in 15 minutes. The results are shown in the table. Name Number of lengths Enzo 11.25 Rashid 18.75 Blessy 20 (i) Find the number of lengths Enzo swims as a percentage of the total number of lengths all three people swim. … % [2] (ii) Write the ratio of the number of lengths each person swims in the form Enzo : Rashid : Blessy. Give your answer in its simplest form. … : … : … [2] (iii) Each length of the pool is 25 m. (a) Work out Blessy’s average swimming speed for the 15 minutes. Give your answer in metres per second. … m/s [3] (b) Rashid continues to swim at the same rate. Calculate the time it takes Rashid to swim a total distance of 5 km. Give your answer in hours and minutes. … h … min [4] (iv) Blessy swims for one hour. The number of lengths she swims decreases by 5% every 15 minutes. Calculate the number of lengths she swims in the final 15 minutes. … [3] (b) Another swimmer, Adam, swims 450 m, correct to the nearest 25 metres. This takes 10 minutes, correct to the nearest minute. Calculate the minimum distance Adam swims in one hour at this rate. … m [3]

17 marks

Mark scheme: 4(a)(i) 22.5 2 11.25 M1 for 100  oe 11.25 + 18.75 + 20 4(a)(ii) 9 : 15 : 16 2 M1 for 1125 : 1850 : 2000 or better 4(a)(iii)(a) 5 3 or 0.556 or 0.5555 to 0.5556 9 20  25 M2 for oe 15[ 60] or M1 for 20 × 25 or for their distance ÷ (15 [× 60]) oe 4(a)(iii)(b) 2 h 40 mins 4 Approach 1 8 B3 for [h]oe or 160 [mins] or 9600[s] 3 Or M3 for 5000 ÷ (18.75 × 25 × 4)[h] oe or 5000 ÷ (18.75 × 25 ÷ 15)[mins] oe or 5000 ÷ ((18.75 × 25 × 4) ÷ (60 × 60))[secs] oe Or M2 for (18.75 × 25 × 4)[m/h] oe or (18.75 × 25 ÷ 15)[m/min] oe or (18.75 × 25 × 4) ÷ (60 × 60))[m/sec] oe Or B1 for 200 or 1 km =1000m soi After 0 scored SC1 for time Figs 267 or figs 2666 to 2667 or figs 16 or figs 96 Approach 2 B3 for 160 [mins] Or M3 for 15 × 5000 ÷ (18.75 × 25) [mins] oe Or M2 for 5000 ÷ (18.75 × 25) oe Or B1 for 200 or 1 km =1000m soi After 0 scored SC1 for time figs 16 4(a)(iv) 17.1 or 17.14 to 17.15 3 3  100 − 5  M2 for 20 ×   oe  100   100 − 5  k or M1 for 20 ×   where k is 2, or  100  4  100 − 5  3 or for 20 ×   oe seen and spoiled  100  4(b) 2500 3 425to450 450 − 12.5 M2 for or or 10 + 0.5 10 to 11 425 to 450 450 − 12.5 or 630 600 to 660 or M1 for 10.5 or 9.5 or 437.5 or 462.5 or 630[s] or 570[s]

This question in 0980/41 Oct/Nov 2024

11 In a sale, the original price of a shirt is reduced by 15%. The sale price of the shirt is $23.63 . Find the original price of the shirt. $ … [2]

2 marks

Mark scheme: 11 27.8[0] 2  M1 for X ×  100 − 15 = 23.63 oe  100 

This question in 0980/42 May/June 2025

17 (a) Alex invests $400 at a rate of 2.3% per year simple interest. Find the total amount Alex has at the end of 5 years. $ … [3] (b) Virat has $100 to spend. In February he spends $x . In March he spends 10% more than he spends in February. In April he spends 10% more than he spends in March. At the end of April, Virat has $33.80 remaining. Find the value of x. x = … [3] (c) Bobbie invests $500 in an account that pays compound interest each year. At the end of 17 years, the value of Bobbie’s investment is $700.13 . Find the value of Bobbie’s investment at the end of 20 years. $ … [4]

10 marks

Mark scheme: 17(a) 446 3 B2 for answer 46 400  2.3  5 or M2 for 400 + oe 100 400  2.3 5  or M1 for oe 100 17(b) 20 nfww 3 M2 for x + 1.1x + 1.12x = [100 –] 33.80 oe  10  2 oe seen or M1 for  1 +  x  100  or for one correctly evaluated trial 17(c) 742.97 to 742.99 4 B3 for 1.02[0…] or interest rate = 2[.0…][%] OR  700.13  20 M3 for 500  17  oe    500   700.13  3 or for 700.13×  17  oe    500  700.13 or M2 for 17 oe 500 OR M1 for 500(…)17 = 700.13 oe M1 dep on previous M1 for  their r  20  their r 3 500  1 + or 700.13  1 +      100   100 

This question in 0980/42 May/June 2025

14 A cube contains a solid metal sphere. The sphere touches all the faces of the cube. The side length of the cube is 8 cm. 256 (a) Show that the volume of the sphere is rcm 3. 3 [1] (b) Calculate the percentage of the cube that is not occupied by the sphere. … % [3] (c) The density of the metal of the sphere is 7.86 g/cm3. Calculate the mass of the sphere. Give your answer in kilograms. [Density = mass ' volume] … kg [2] (d) The sphere is melted down and made into a solid cylinder with radius 3.1 cm. Calculate the total surface area of the cylinder. … cm2 [4]

10 marks

Mark scheme: 14(a) 4 3 256 1 π 4 [= π ] 3 3 14(b) 47.6 nfww or 3 50 B2 for 52.4 or 52.35 to 52.37 or π nfww 47.63 to 47.64… nfww 3 OR 3 256 8 − π 3 M2 for 3  100 oe 8 3 256 256 8 − π π 3 3 or M1 for 3 [ 100] oe or 3  100 8 8 oe 14(c) 2.11 or 2.107… 2 256 M1 for π  7.86 3 14(d) 233 or 233.3 to 233.4 4 2 256 M1 for π  3.1  h = π 3 M2dep for 2  π  3.12 + 2  π  3.1  their h or M1dep for 2 π 3.1  theirh or M1 for 2π  3.12

This question in 0980/41 Oct/Nov 2025